You measure 21 watermelons' weights, and find they have a mean weight of 71 ounces. Assume the population standard deviation is 3 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Give your answers as decimals, to two places

To construct a 99% confidence interval for the true population mean, we can use the formula for the confidence interval for the population mean when the population standard deviation \(\sigma\) is known:

\[
\text{Confidence Interval} = \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]

Where:
- \(\bar{x}\) is the sample mean (71 ounces),
- \(z_{\alpha/2}\) is the z-score corresponding to the desired confidence level (for a 99% confidence level, \(z_{\alpha/2} = 2.576\)),
- \(\sigma\) is the population standard deviation (3 ounces),
- \(n\) is the sample size (21).

Now, let's calculate the margin of error:

\[
\text{Margin of Error} = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} = 2.576 \cdot \frac{3}{\sqrt{21}}
\]

Let's calculate that.The margin of error is approximately 1.69 ounces.

Now, we can calculate the 99% confidence interval:

\[
\text{Confidence Interval} = \bar{x} \pm \text{Margin of Error}
\]

This gives us:
\[
71 \pm 1.69
\]

So, the 99% confidence interval is:

\[
(71 - 1.69, 71 + 1.69) = (69.31, 72.69)
\]

Thus, the 99% confidence interval for the true population mean watermelon weight is \((69.31, 72.69)\) ounces.

Would you like more details, or do you have any questions? Here are 5 related questions:

1. How does the confidence level affect the width of the confidence interval?
2. What would the confidence interval be if the confidence level were 95%?
3. How would the confidence interval change if the sample size were larger?
4. Can you explain why we use the z-score instead of the t-score in this case?
5. What assumptions are made when constructing this confidence interval?

**Tip:** A higher confidence level results in a wider interval, as it aims to capture the true population mean with greater certainty.

You measure 21 watermelons' weights, and find they have a mean weight of 71 ounces. Assume the population standard deviation is 3 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Learn how to construct a 99% confidence interval for the true population mean weight of watermelons using a sample size of 21, a mean weight of 71 ounces, and a population standard deviation of 3 ounces. This problem covers key statistical concepts, including confidence intervals, z-scores, and the central limit theorem. Suitable for students in Grades 11-12 or introductory statistics courses.

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